Question

A solid cylinder of mass 4Kg and radius 4cm is rotating about its axis at the rate of 3rpm. What is the torque required to stop after $ 2\pi $ revolutions?
(A) $ 2\times {{10}^{-3}}Nm $
(B) $ 12\times {{10}^{-4}}Nm $
(C) $ 2\times {{10}^{6}}Nm $
(D) $ 2\times {{10}^{-6}}Nm $

Answer 1

Hint: Torque is the rotational equivalent of force. So, a net torque will cause an object to rotate or stop its rotation with an angular acceleration. Because all rotational motions have an axis of rotation, a torque must be defined about a certain rotational axis.
A torque is basically a force applied to a point on an object about the axis of rotation. The size of a torque depends on the amount of the force applied and its perpendicular distance from the axis of rotation (which depends both on the direction of the force and its physical distance from the axis of rotation as it is a vector quantity).
Torque required to stop a rotating body is given by
  $ \tau =\iota \alpha $
Where $ \tau $ is the torque, $ \iota $ is the moment of inertia of the body about the given axis and $ \alpha $ is the angular acceleration of the body.

Complete step by step solution:
We are given:
Angular velocity of the solid cylinder; $ {{\omega }_{0}}=3rpm $
$ \begin{align}
& {{\omega }_{0}}=3\times \dfrac{2\pi }{60}rad/s \\
& {{\omega }_{0}}=\dfrac{\pi }{10}rad/s \\
\end{align} $
As the cylinder is stopped at the end so final angular velocity is zero.
Angular distance travelled after which the cylinder is to be stopped;
$ \begin{align}
& \theta =2\pi \times 2\pi \\
& \theta =4{{\pi }^{2}} \\
\end{align} $
( $ \because $ It covers $ 2\pi $ distance $ 2\pi $ times)
Using Newton’s third equation of motion in angular terms i.e.
$ {{\omega }^{2}}={{\omega }_{0}}^{2}+2\alpha \theta $
So,
$ \begin{align}
& 0={{\left( \dfrac{\pi }{10} \right)}^{2}}+2\alpha \left( 4{{\pi }^{2}} \right) \\
& -\dfrac{{{\pi }^{2}}}{100}=8{{\pi }^{2}}\alpha \\
\end{align} $
$ \alpha =-\dfrac{1}{800}rad/{{s}^{2}} $
The moment of inertia of a solid cylinder considering it to be rotating about its central axis is given as
$ \iota =\dfrac{1}{2}m{{R}^{2}} $
$ \begin{align}
& \iota =\dfrac{1}{2}\left( 2 \right){{\left( \dfrac{4}{100} \right)}^{2}} \\
& \\
& \iota =16\times {{10}^{-4}}Kg{{m}^{2}} \\
\end{align} $
Now calculating the torque:
$ \begin{align}
& \tau =\iota \alpha \\
& \tau =16\times {{10}^{-4}}\times \left( -\dfrac{1}{800} \right) \\
& \tau =-2\times {{10}^{-6}}Nm \\
\end{align} $
Here negative sign in torque implies that it is stopping the rotating cylinder from its motion.
Therefore, option (D) is the correct answer.

Note:
We must never ignore the fact that one revolution sweeps the angular distance of $ 2\pi $ and in this question, we are given $ 2\pi $ revolutions which means the total distance swept is $ \left( 2\pi \times 2\pi \right) $ . This is the major mistake which mostly occurs when instead of considering $ \theta =2\pi \times 2\pi $ , we consider only $ 2\pi $ giving us the wrong answer.